Find the slope and y-intercept of the line that is ${\text{perpendicular}}$ to $\enspace {y = -3x }\enspace$ and passes through the point ${(6, 8)}$. ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$ ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$
Answer: Lines are considered perpendicular if their slopes are negative reciprocals of each other. The slope of the blue line is ${-3}$ , and its negative reciprocal is ${\dfrac{1}{3}}$ Thus, the equation of our perpendicular line will be of the form $\enspace {y = \dfrac{1}{3}x + b}\enspace$ We can plug our point, $(6, 8)$ , into this equation to solve for ${b}$ , the y-intercept. $8 = {\dfrac{1}{3}}(6) + {b}$ $8 = 2 + {b}$ $8 - 2 = {b} = 6$ The equation of the perpendicular line is $\enspace {y = \dfrac{1}{3}x + 6}\enspace$. ${m = \dfrac{1}{3}, \enspace b = 6}$